Help with Math Proof

[Scrapster]

Problem:

Let A be the set of integers expressible as (2k - 1) for some k that "belongs to" Z (set of integers). Let B be the set of integers expressible as (2k + 1) for some k that "belongs to" Z (set of integers). Prove that A=B.

I'm stuck on how to start this. Anyone have ideas?

 

[hendon]

You can show a one-one correspondence between the two sets
or something like

2k + 1 = 2(k+1) - 1
where k is an integer

ie every number that belongs to the first set belongs to the second set
and vice versa
then A=B

 

[NikPreviousAcct]

<edit>

oh... nevermind.

</edit>

 

[BlueScreenVW]

You realize that the problem depends on k being able to go to infinity, right? Anyway, I believe that this problem is almost more of a logical than a mathematical type. Use the letter n instead of one of the k:s, and you get (with a being any number in A, and b any in B:

a=b <--> 2k-1=2n+1 <--> 2k=2n+2 <--> k=n+1, and since n+1 is also an integer k still belongs to Z. That is, for any number a in A there exists a number b in B, and vice versa (btw, don't forget that vice versa when it comes to more complicated problems).

Note: Compare this to a problem where A is the even numbers (2k), and B is the odd (2n+1), where you'll get like k=n+1/2 in the end - here it's obvious that both k and n can not be integers.

 

[Scrapster]

I think our goal is to show that A and B are subsets of each other.

 

[StormRider]

The general approach to show that two sets A and B are equal is to show that every element x in A is also in B and that every element y in B is also in A. This shows that A is a subset of B and that B is a subset of A and this implies that A = B.

So, you start off with assuming that x is any element in A.
Since x is an element in A, it can be express as x = 2k - 1 for some integer k.

Now we need to show that x is also an element in B.

Let's rewrite x with some simple arithmetic operations.

x = 2k - 1
= 2k - 1 - 1 + 1
= 2k - 2 + 1
= 2(k-1) + 1
x = 2m + 1 where m = k-1 is some integer

Notice that x is of the same form as elements of the set B? Therefore x is an element of B.

Since x was any arbitrary element of A, this implies that A is a subset of B.

Now, we have to show that B is a subset of A. The method is exactly analogous to showing A is a subset of B so I'll leave the rest to you...

 

[hendon]

For every element x belonging to A,
x = 2k-1 for some integer k
x = 2(k-2)+1 for some integer k
since k-2 is also an integer, x belongs to B
therefore, A is a subset of B

then do a similar thing for the reverse to show that B is a subset of A

then conclude A=B

 

[Scrapster]

I got the idea. Thanks guys.

This is my first time using sets with proofs. Is it pretty common to use substitution and sticking in +1's and -1's? I didn't even think about that. But it definately makes sense in this case.

 

[Scrapster]

I got the idea. Thanks guys.

This is my first time using sets with proofs. Is it pretty common to use substitution and sticking in +1's and -1's? I didn't even think about that. But it definately makes sense in this case.